Finding limits by algebra

What you will be able to do

• turn $\dfrac{0}{0}$ into a finite limit by factorising or by rationalising;
• find a limit as $x \to \infty$ by dividing by the highest power;
• read the limit of a fraction of polynomials from the degrees.

When substitution fails

You try $x = a$ and the answer is $\dfrac{0}{0}$. Do not stop. A limit ignores what happens exactly at $a$. So we replace the fraction by another one that is the same for every $x \ne a$ and that can be evaluated at $a$.

The method has three steps.

1. Substitute. If you get a number, you are done.
2. If you get $\dfrac{0}{0}$, find the hidden factor $(x - a)$. Factorise, or rationalise.
3. Cancel the factor, then substitute again.

Factorise and cancel

Find $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 5x + 6}{x - 2}$.

At $x = 2$ we get $\dfrac{0}{0}$. The top factorises as $(x - 2)(x - 3)$. For $x \ne 2$ the fraction is $x - 3$, and its limit at $2$ is $-1$.

A cube gives the same pattern. For $\displaystyle\lim_{x \to 2} \dfrac{x^3 - 8}{x - 2}$ use $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$. After cancelling, put $x = 2$ in $x^2 + 2x + 4$ to get $12$.

Rationalise a square root

A square root in the top hides the factor. Multiply the top and the bottom by the conjugate, which is the same expression with the sign between the terms changed.

Find $\displaystyle\lim_{x \to 0} \dfrac{\sqrt{1 + x} - 1}{x}$.

1. Multiply by $\dfrac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}$. The top becomes $(1 + x) - 1 = x$.
2. The fraction is now $\dfrac{x}{x(\sqrt{1 + x} + 1)} = \dfrac{1}{\sqrt{1 + x} + 1}$ for $x \ne 0$.
3. At $x = 0$ this is $\dfrac{1}{2}$.

So the limit is $\dfrac{1}{2}$. This is also the slope of $\sqrt{1 + x}$ at $x = 0$, a number you will meet again in the derivative.

Limits at infinity

Now let $x$ grow without bound. We write $x \to \infty$. The key fact is that $\dfrac{1}{x} \to 0$, and so $\dfrac{1}{x^2} \to 0$ as well.

To find the limit of a fraction of polynomials, divide the top and the bottom by the highest power of $x$.

As $x \to \infty$, the terms $\dfrac{5}{x}$ and $\dfrac{1}{x^2}$ tend to $0$. So the limit is $\dfrac{3}{2}$.

The graph settles near the horizontal line $y = \dfrac{3}{2}$. That line is a horizontal asymptote.

Comparing the degrees gives a quick rule for $\dfrac{P(x)}{Q(x)}$:

• if the degree of $P$ is less than the degree of $Q$, the limit is $0$;
• if the degrees are equal, the limit is the ratio of the leading coefficients;
• if the degree of $P$ is greater, the values grow without bound and there is no finite limit.

A square root at infinity

Find $\displaystyle\lim_{x \to \infty} \big(\sqrt{x^2 + x} - x\big)$. Each part grows without bound, so we cannot subtract the limits. Rationalise instead.

1. Multiply by $\dfrac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}$. The top becomes $x^2 + x - x^2 = x$.
2. The expression is $\dfrac{x}{\sqrt{x^2 + x} + x}$. Divide the top and the bottom by $x$ (for $x > 0$).
3. This gives $\dfrac{1}{\sqrt{1 + \frac{1}{x}} + 1}$, whose limit is $\dfrac{1}{2}$.

The answer is $\dfrac{1}{2}$, not $0$. The two big numbers are not equal enough to cancel. This is called an $\infty - \infty$ form, and course 3 comes back to it.

Practise with new numbers

Try it in Python

The program evaluates $\dfrac{3x^2 + 5x}{2x^2 - 1}$ at $x = 10, 100, \dots$ and then the difference $\sqrt{x^2 + x} - x$ in its safe form.

Common mistakes

• Cancelling terms, not factors. In $\dfrac{x + 3}{x + 1}$ you may not cancel the $x$. Only a factor of the whole top and the whole bottom can go.
• Using $\infty$ as a number. $\dfrac{\infty}{\infty}$ is not $1$. Divide by the highest power instead.
• Forgetting the conjugate on both lines. Multiply the top and the bottom by the same expression.
• Subtracting two infinite limits. $\infty - \infty$ is not $0$. Rationalise first.

Remember these

• Substitute first. If you get $\dfrac{0}{0}$, factorise or rationalise, cancel, and substitute again.
• As $x \to \infty$, divide the top and the bottom by the highest power of $x$.
• Fraction of polynomials: lower degree on top gives $0$, equal degrees give the ratio of leading coefficients, higher degree on top gives no finite limit.
• $\dfrac{1}{x} \to 0$ as $x \to \infty$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x^2 - 5x + 6}$.

2. Find $\displaystyle\lim_{x \to 0} \dfrac{\sqrt{4 + x} - 2}{x}$.

3. Find $\displaystyle\lim_{x \to \infty} \dfrac{4x^3 - x}{2x^3 + 7x^2}$.

4. Find $\displaystyle\lim_{x \to \infty} \dfrac{5x + 1}{x^2 + 3}$.

5. Find $\displaystyle\lim_{x \to 1} \dfrac{x^3 - 1}{x^2 - 1}$.

References

• OpenStax, Calculus Volume 1, Chapter 2, section 2.3 on the limit laws, and Chapter 4, section 4.6 on limits at infinity and asymptotes.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the sections on limits of polynomials and rational functions.